Math 10120 — Spring 2013
Some more practice problems for the final
Instructor: David Galvin
1. Of the 75 students in Math 10120, 45 are women. Once during each lecture, I pick a student
at random to ask a question to. There are 41 lectures during the semester. Use the normal
approximation to the binomial to estimate the probability that during the course of the
semester, I pick at least 30 men to ask questions to?
Solution: If X is the number of men I pick, then X is binomial with n = 41 and p = 30/75 =
√
.4. We have µ = E(X) = np = 16.4 and σ = npq ≈ 3.14. We want Pr(X ≥ 30). Using
the normal approximation, we calculate Pr(X ≥ 29.5) where X is normal with mean 16.4,
standard deviation 3.14. Normalizing to a standard normal, this is the same as Pr(Z ≥ 4.18).
This is (basically) 0.
2. When I throw a discus, my average distance is 50 meters, with standard deviation 5. My
throw distances are definitely not normally distributed, but still, from the information given,
you can say things for certain. For example: With 90% probability, my throws will always be
between 50 − a meters and 50 + a meters. What is a?
Solution: Using Tchebychev, the probability of a throw being in the range 50 − a meters to
50 + a meters is at least 1 − (σ 2 /a2 ) = 1 − (25/a2 ). To make this .9, we need 25/a2 = .1, or
a2 = 250, or a ≈ 15.8.
3. I want to determine something about the speeds at which cars drive past the pedestrian
crossing at the corner of Vaness and Twyckenham. I take measurements of 10 cars, and get
the following readings from my speed gun (all measurements in miles per hour):
25, 35, 45, 50, 55, 35, 30, 45, 45, 35.
Find
(a) the mean of the sample data;
Solution: x = 25+35+45+50+55+35+30+45+45+35
= 40.
10
(b) the sample variance of the sample;
Solution:
2
2
2
2
2
2 +(30−40)2 +(45−40)2 +(45−40)2 +(35−40)2
s2 = (25−40) +(35−40) +(45−40) +(50−40) +(55−40) +(35−40)
=
9
88.88 . . ..
(c) the sample standard
deviation of the data.
√
2
Solution: s = s = 9.428 . . ..
4. A confectioner makes two raisin-nut mixtures. A Standard box contains 6oz of peanuts, 1oz
of raisins and 4oz of cashews, and sells for $4.25. A Deluxe box contains 12oz of peanuts,
3oz of raisins and 2oz of cashews, and sells for $6.55. The confectioner currently has in stock
5400oz of peanuts, 1200oz of raisins, and 2400oz of cashews.
1
(a) Write down the linear programming problem that the confectioner has to solve, to decide how many Standard and how many Deluxe boxes to make, to maximize revenue
(assuming he sells all boxes he makes).
Solution: Let S be number of Standard boxes made, and D the number of Deluxe.
The LP is to maximize 4.25S + 6.55D, subject to 6S + 12D ≤ 5400, S + 3D ≤ 1200,
4S + 2D ≤ 2400 and S, D ≥ 0.
(b) Show the feasible set on a graph.
Solution:
The feasible set is the 5-sided shape with corners A, B, C, D and E shown in the picture
below (x-axis is S, y-axis is D):
(c) Find the corners of the feasible set.
Solution:
i.
ii.
iii.
iv.
v.
A : (0, 0) [Objective value $0]
B : (0, 400) [Objective value $2, 620]
C : (300, 300) [Objective value $3, 240]
D : (500, 200) [Objective value $3, 435]
E : (600, 0) [Objective value $2, 550]
(d) Solve the linear programming problem.
Solution: From the calculations of the objective function at each corner of the feasible
set, done in the last part, we see that the optimal solution is to make 500 boxes of
Standard and 200 boxes of Deluxe (corner D).
(e) What is the confectioners optimal profit?
Solution: $3, 435.
(f) If the confectioner makes the mix of Standard and Deluxe boxes dictated by the solution
to the linear programming problem, what inventory does he have left after he has made
all the boxes?
2
Solution: At S = 500 and D = 200, he uses 5400oz of peanuts (so has none left), 1100oz
of raisins (so has 100oz left) and 2400oz of cashews (so has none left). So: 100oz of raisins
is all that is left.
3